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A basket has 13 black balls, 12 white balls, and 15 Red balls. If 4 balls are picked at random then find the probability of getting 2 black balls and 2 red balls.1. 79/8132. 60/7033. 63/7034. 703/63 |
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Answer» Correct Answer - Option 3 : 63/703 Given: Black balls = 13 White balls = 12 Red balls = 15 Formula used: nCr = n!/((n-r)! × r!) Calculation: Let S be the sample space then, n(S) = Number of ways of selecting 4 balls out of 40 ball ⇒ n(S) = 40C4 = 40!/((40 – 4)! × 4!) ⇒ (40!/36! × 4!) = (40 × 39 × 38 × 37)/(4 × 3 × 2) Let E be the event of getting 2 black and 2 red ball n(E) = 13C2 × 15C2 ⇒ (13 × 12)/2 × (15 × 14)/2 P(E) = n(E)/n(S) ⇒ P(E) = \(\frac{{\frac{{13{\rm{\;}} × {\rm{\;}}12}}{2} × \frac{{{\rm{\;}}\left( {15{\rm{\;}} × {\rm{\;}}14} \right)}}{2}}}{{\frac{{40{\rm{\;}} × {\rm{\;}}39{\rm{\;}} × {\rm{\;}}38{\rm{\;}} × {\rm{\;}}37}}{{4{\rm{\;}} × {\rm{\;}}3{\rm{\;}} × {\rm{\;}}2}}}}\) ⇒ (13 × 12 × 15 ×14 × 3 × 2)/(40 × 39 × 38 × 37) ⇒ 63/703 ∴ The probability of getting 2 black ball and 2 red ball is 63/703 |
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